NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Maths Solutions Chapter 5: Exercise - 1.3

Question:1

For a positive integer n, find the value of $(1 - i)^n$ $(1-1/i)^n$

Answer:

$(1-i)^n (1-1/i)^n$
$=(1-i)^n (1+i )^n$
$=(1-i^2 )^n$
$=2^n$

Question:2

Evaluate $\sum_{i=1}^{13}$$(i^n+i^{n+1} )$ where n $\epsilon$N.

Answer:

= $\sum_{i=1}^{13}$$(i^n+i^{n+1} )$
$=$ $\sum_{i=1}^{13}$$(1+i) i^n$
$=(1+i)(1+i^2+i^3+i^4....+i^{12}+i^{13} )$
$=(1+i) \frac{i(i^{13}-1)}{(i-1)}$
$=(1+i) i\frac{(i-1)}{(i-1)}$
$=(1+i)i$
$=i+i^2$
$=i-1$

Question:3

If $\frac{(1+i)}{(1-i)}^3-\frac{(1-i)}{(1+i)}^3= x+iy$then find (x, y).

Answer:

Given $\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3= x+iy$
$=\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3$
$=\left (\frac{1+2i+i^2}{1-i^2} \right )^{3}-\left (\frac{1-2i+i^2}{1-i^2} \right )^{3}$
$=(\frac{2i}{2})^3-(\frac{-2i}{2})^3$
$=i^3-(-i^3 )$
$=2i^3=0-2i$
$Thus,(x,y)=(0,-2)$

Question:4

If $(1+i)^2/(2-i)= x+iy$ then find the value of x + y.

Answer:

$x+iy=\frac{(1+i)^2}{(2-i)}=\frac{ (1+2i+i^2)}{(2-i)}= \frac{2i}{(2-i)}$

$On\: \: rationalising\: \: 2i(2+i)/(2-i)(2+i)$,
$=\frac{(4i+2i^2)}{(4-i^2 )}=\left (\frac{(4i-2)}{(4+1)} \right )$
$=\frac{-2}{5}+\frac{4i}{5}$
$x=\frac{-2}{5} \: \: and\: \: y=\frac{4}{5}$

$So, x+y= -\frac{2}{5}+\frac{4}{5}=\frac{2}{5}$

Question:5

If $\left(\frac{1-i}{1+i}\right)^{100}=a+ib$ then find (a, b).

Answer:

$a+ib=\left(\frac{1-i}{1+i}\right)^{100}=\left [\frac{1-i}{1+i}*\frac{1-i}{1-i} \right ]^{100}$
$=\left [\frac{(1-i)^2}{1-i^2} \right ]^{100}$
$=\left (\frac{(1-2i+i^2)}{(1+1)} \right )^{100}$
$=\left (\frac {-2i}{2 }\right )^{100}$
$=(i^4 )^{25}=1$
$Hence,(a,b)=(1,0)$

Question:6

If $a=\cos \theta+i \sin\theta$, find the value of$\frac{(1+a)}{(1-a)}$ .

Answer:

$a=\cos \theta+i \sin\theta$
$\frac{(1+a)}{(1-a)}$$=\frac{(1+cos\theta)+i \sin \theta}{(1-cos\theta)-isin\theta}$

$=\frac{\left ( 2\cos ^{2}\frac{\theta }{2} +i2\sin\frac{\theta }{2}\cos\frac{\theta }{2} \right )}{\left ( 2\sin ^{2}\frac{\theta }{2} -i2\sin\frac{\theta }{2}\cos\frac{\theta }{2} \right )}$
$=\frac{2\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{2\sin\frac{\theta }{2}\left ( \sin\frac{\theta }{2}-i\cos\frac{\theta }{2} \right )}$
$=\frac{i\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{\sin\frac{\theta }{2}\left ( i\sin\frac{\theta }{2}-i^2\cos\frac{\theta }{2} \right )}$
$=\frac{i\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{\sin\frac{\theta }{2}\left ( i\sin\frac{\theta }{2}+\cos\frac{\theta }{2} \right )}$
$=i\cot \frac{\theta }{2}$

Question:7

If $(1 + i)z = (1 - i) \bar{z},$ then show that $z= -i\bar{z}$

Answer:

$(1 + i)z = (1 - i) \bar{z},$
$z=\frac{1-i}{1+i}\bar{z}$
Rationalising the denominator,
$\frac{\left (1-i \right )\left (1-i \right )}{\left (1+i \right )\left (1-i \right )}\bar{z}$
$=\frac{\left ( 1-i \right )^{2}}{1-i^2}\bar{z}$
$=\frac{\left ( 1-2i+i^2 \right )}{1+1}\bar{z}$
$=\frac{\left ( 1-2i-1 \right )}{2}\bar{z}$
$=-i\bar{z}$
Hence,Proved.

Question:8

If$z = x + iy$, then show that $z\bar{z}+2(z+\bar{z})+b=0$ where b?R, representing z in the complex plane is a circle.

Answer:

$z=x+iy$
$\bar{z}=x-iy$
$Now,we\: \: also\: \: have\: \: ,z\bar{z }+2(z+\bar{z})+b=0$
$(x+iy)(x-iy)+2(x+iy+x-iy)+b=0$
This equation is a equation of circle

Question:9

If the real part of $\frac{(\bar{z}+2)}{(\bar{z}-1)}$ is 4, then show that the locus of the point representing z in the complex plane is a circle.

Answer:

$Let \: \: z=x+iy \: \:$
$\frac{\bar{z}+2}{\bar{z}-1}= \frac{x-iy+2}{x-iy-1}$
$=\frac{\left [ \left ( x+2 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}{\left [ \left ( x-1 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}$
$=\frac{(x-1)(x+2)+y^2+i[(x+2)y-(x-1)y]}{(x-1)^2+y^2}$
$real \: \: part=4 \Rightarrow \frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4$
$x^2+x-2+y^2=4(x^2-2x+1+y^2 )$
$3x^2+3y^2-9x+6=0$

The equation obtained is that of a circle. Hence, locus of z is a circle.

Question:10

Show that the complex number z, satisfying the condition $arg\left ( \frac{z-1}{z+1} \right )=\frac{\pi}{4}$ lies on a circle.

Answer:

Let z=x+iy
$arg\left ( \frac{z-1}{z+1} \right )=\frac{\pi}{4}$
$arg\left ( z-1 \right )-arg\left ( z+1 \right )=$$\frac{\pi}{4}$
$arg\left ( x+iy-1 \right )-arg\left ( x+iy+1 \right )=$$\frac{\pi}{4}$
$arg\left ( x-1+iy \right )-arg\left ( x+1+iy \right )=$$\frac{\pi}{4}$
$\tan^{-1}\frac{y}{x-1}-\tan^{-1}\frac{y}{x+1}=$$\frac{\pi}{4}$
$\tan^{-1}\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\left (\frac{y}{x-1} \right )\left (\frac{y}{x+1} \right )}=$$\frac{\pi}{4}$
$\tan^{-1}\left(\frac{y\left ( x+1+-x+1 \right )}{x^2-1+y^2}\right) = \frac{\pi}{4}$
$\tan\frac{\pi}{4} = \frac{2y}{x^2+y^2-1}$
$x^2+y^2-1=2y$
$x^2+y^2-1-2y=0$
The equation obtained represents the equation of a circle

Question:11

Solve that equation $|z| = z + 1 + 2i$.

Answer:

$|z| = z + 1 + 2i$
$Substituting \: \: z=x+iy \: \: we \: \: get \: \: |x+iy|=x+iy+1+2i$
$|z|= \sqrt{(x^2+y^2 )} =(x+1)+i(y+2)$
Comparing real and imaginary parts $\sqrt{(x^2+y^2 )} =(x+1)$
And$0=y+2 , y=-2$
Substituting the value of y in$\sqrt{(x^2+y^2 )} =(x+1)$
$x^2+(-2)^2=(x+1)^2$
$x^2+4=x^2+2x+1$

Hence, $x=\frac{3}{2}$
Hence,$z=x+iy$
$=\frac{3}{2} -2i$

Question:12

If $|z + 1| = z + 2 (1 + i)$, then find z

Answer:

We have $|z + 1| = z + 2 (1 + i)$
Substituting $z=x+iy$ we get $x+iy+1=x+iy+2i+1$
$|z|= \sqrt{(x^2+y^2 ) }=\sqrt{(x+1)^2+y^2 } = (x+2)+i(y+2)$
Comparing real and imaginary parts, $\sqrt{(x+1)^2+y^2 } = (x+2)$
$and 0=y+2 ;y=-2$
Substituting the value of y in $\sqrt{(x+1)^2+y^2 } = (x+2)$
$(x+1)^2+(-2)^2=(x+2)^2$
$x^2+2x+1+4=x^2+4x+4$
$2x=1$
Hence, $x=\frac{1}{2}$
Thus, $z=x+iy=\frac{1}{2}-2i$

Question:13

If $arg (z - 1) = arg (z + 3i)$, then find x – 1 : y. where $z = x + iy$

Answer:

Given that $arg (z - 1) = arg (z + 3i)$
$arg (x+iy-1)=arg(x+iy+3i)$
$arg(x-1+iy)=arg(x+i(y+3))$
$\tan^{-1}\frac{y}{x-1}=\tan^{-1}\frac{y+3}{x}$
$\frac{y}{x-1}=\frac{y+3}{x}$
$xy=xy-y+3x-3$
$3x-3=y$
$\frac{x-1}{y}=\frac{1}{3}$

Question:14

Show that $\left | \frac{z-2}{z-3} \right |=2$ represents a circle. Find its centre and radius.

Answer:

$\left | \frac{z-2}{z-3} \right |=2$ Substituting $z=x+iy$, we get $\left | \frac{x+iy-2}{x+iy-3} \right |=2$
$|x-2+iy|=2|x-3+iy|$
$\sqrt{(x-2)^2+y^2 }=2\sqrt{((x-3)^2+y^2}$
$x^2-4x+4+y^2=4(x^2-6x+9+y^2 )$
$3x^2+3y^2-20x+32=0$
$x^2+y^2-\frac{20}{3} x+\frac{32}{}3=0$
$(x-\frac{10}{3})^2+y^2+\frac{32}{3}-\frac{100}{9}=0$
Thus, the centre of circle is $(\frac{10}{3},0)$ and radius is $2/3$

Question:15

If $\frac{z-1}{z+1}$ is a purely imaginary number$(z \neq -1)$, then find the value of $|z|$.

Answer:

Let $z=x+iy$
$\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}$
$\frac{\left [ \left ( x-1 \right ) +iy\right ]\left [ \left ( x+1 \right )-iy \right ]}{\left [ \left ( x+1 \right ) +iy\right ]\left [ \left ( x+1 \right )-iy \right ]}$
$=\frac{(x-1)(x+1)+y^2+i[(x+1)y-(x-1)y]}{(x+1)^2+y^2 }$
$\frac{z-1}{z-1}$ is purely imaginary
$\frac{(x-1)(x+1)+y^2}{(x+1)^2+y^2 }=0$
$x^2-1+y^2=0$
$x^2+y^2=1$ Hence, $|z|=1$

Question:16

z₁ and z₂ are two complex numbers such that $|z_1| = |z_2|$ and $arg (z_1) + arg (z_2) = \pi$, then show that $|z_1| = -|z_2|$ .

Answer:

Let $z_1=\left |z_1 \right |\left (cos\theta _1+i\ sin\theta_1 \right )$ and $z_2=\left |z_2 \right |\left (cos\theta _2+i\ sin\theta_2 \right )$
Given that $|z_1| = |z_2|$
And $arg\left (z_1 \right )+arg\left (z_2 \right )= \pi$

$\theta_1+\theta_2=\pi$

$\theta_1=\pi-\theta_2$

$z_2=\left |z_2 \right |\left (cos\theta _2+i\ sin\theta_2 \right )$

$z_1=|z_2 |\left ( -\cos \theta_2+isin\theta_2 \right )$

$z_1=-|z_2 |\left ( \cos \theta_2-isin\theta_2 \right )$

$z_1=-|z_2 |\left ( \cos \theta_2-isin\theta_2 \right )$, $|z_1| = -|z_2|$

Question:17

If |z₁| = 1 (z₁ ≠ –1) and $z_2= \frac{z_1-1}{z_1+1}$ then show that the real part of z₂ is zero

Answer:

Let $z_1=x+iy$ $|z_1 |= \sqrt{x^2+y^2} =1$
$z_2= \frac{z_1-1}{z_1+1}=\frac{x+iy-1}{x+iy+1}$
$=\frac{[(x-1)+iy][(x+1)-iy]}{[(x+1)+iy][(x+1)-iy] }$
$= \frac{(x-1)(x+1)+y^2+i[(x+1)y-(x-1)y]}{(x+1)^2+y^2}$
$\frac{x^2+y^2-1+2iy}{(x+1)^2+y^2 }=0$
Since, $x^2+y^2=1$
$\frac{1-1+2iy}{(x+1)^2+y^2}=\frac{0+2iy}{(x+1)^2+y^2}$
Therefore, the real part of $z_2$ is zero

Question:18

If z₁, z₂ and z₃, z₄ are two pairs of conjugate complex numbers, then find $arg\left ( \frac{z_1}{z_4} \right )+arg\left ( \frac{z_2}{z_3} \right )$

Answer:

$z_1$ and $z_2$ are conjugate complex numbers.
The negative side of the real axis $= r_1 (cos\theta-isin \theta)$
$=r_1\left ( \cos\left ( -\theta_1 \right )+i\sin\left (- \theta_1 \right ) \right )$
Similarly, $z_3=r_2\left ( \cos\left ( \theta_2 \right )-i\sin\left ( \theta_2 \right ) \right )$
$z_4=r_2\left ( \cos\left ( -\theta_2 \right )+i\sin\left (- \theta_2 \right ) \right )$
$arg \left ( \frac{z_1}{z_4} \right )+arg \left ( \frac{z_2}{z_3} \right )$$=arg(z_1)-arg(z_4)+ arg(z_2)- arg(z_3)??$
$=\theta_1-(-\theta_2)+(-\theta_1)-\theta_2=0$

Question:19

If $|z_1| = |z_2| =... = |z_n| = 1$, then
Show that $|z_1 + z_2 + z_3 +... + z_n| =|1/z_1 +1/z_2 +...+1/z_n |$

Answer:

$|z_1| = |z_2| =... = |z_n| = 1$
$|z_1|^2 = |z_2|^2 =... = |z_n|^2 = 1$
$z_1 \bar{z_1 } =z_2 \bar{z_2 }=...=z_n \bar{z_n }$
$z_1=\frac{1}{\bar{z_1}},z_2=\frac{1}{\bar{z_2}},...z_n=\frac{1}{\bar{z_n}}$
Now, $|z_1+z_2+z_3+...z_n |$ $=\left | \frac{z_1\bar{z_1}}{\bar{z_1}}+\frac{z_2\bar{z_2}}{\bar{z_2}}+...+\frac{z_n\bar{z_n}}{\bar{z_n}} \right |$
=$\left | \frac{1}{\bar{z_1}}+\frac{1}{\bar{z_2}}+...+\frac{1}{\bar{z_n}} \right |$
$=\left | {\frac{1}{z_1}+\frac{1}{z_2}+...+\frac{1}{z_n}} \right |$

Question:20

If for complex numbers z₁ and z₂, $arg (z_1) - arg (z_2) = 0$, then show that $\left | z_1-z_2 \right |= |z_1|-|z_2|$

Answer:

Let $z_1=\left | z_1 \right |\left ( \cos\theta_1 +i\sin \theta_1 \right )$ and $z_2=\left | z_2 \right |\left ( \cos\theta_2 +i\sin \theta_2 \right )$
$arg(z_1 )- arg(z_1 )=0$
$\theta _1-\theta _2=0$
$\theta _1=\theta _2$
$z_1-z_2= (z_1 |cos\theta_1-| z_2|cos\theta_1 )+i(z_1 |sin\theta_1-| z_2|sin\theta_1 )$
$z_1-z_2= \sqrt{(z_1 |cos\theta_1-| z_2|cos\theta_1 )^{2}+(z_1 |sin\theta_1-| z_2|sin\theta_1 )^{2}}$
$=\sqrt{|z_1|^2+|z_2|^2-2|z_1||z_2|\cos^2\theta_1-2|z_1||z_2|\sin^2\theta_1}$
$=\sqrt{|z_1|^2+|z_2|^2-2|z_1||z_2|}$
$=\sqrt{\left (|z_1|-|z_2| \right )^2}= |z_1|-|z_2|$

Question:21

Solve the system of equations $Re (z^2) = 0, |z| = 2.$

Answer:

$Re (z^2) = 0, |z| = 2.$
$Let\: \: z=x+iy$
$|z|= \sqrt{x^2+y^2 } =2$
${x^2+y^2 } =4$
$z^2=x^2+2ixy-y^2$
$=(x^2-y^2 )+2ixy$
$Now,Re (z^2 )=0$
$x^2-y^2=0$
$x^2=y^2=2$
$x=y=\pm \sqrt{2}$
Hence, $z=x+iy=\pm\sqrt{2}\pm i\sqrt{2}$
$=\sqrt{2}+i\sqrt{2},\sqrt{2}-i\sqrt{2},-\sqrt{2}+i\sqrt{2} \: \: and\: \: -\sqrt{2}-i\sqrt{2}$

Question:22

Find the complex number satisfying the equation $z+\sqrt{2} |(z+1)|+i=0$ .

Answer:

$z+\sqrt{2} |(z+1)|+i=0\: \: \: \: ...(i)$
Substituting $z=x+iy$ we get $x+iy+\sqrt2 |x+iy+1|+i=0$
$x+i(1+y)+\sqrt2 \left [ \sqrt{(x+1)^2+y^2 } \right ]=0$
$x+i(1+y)+\sqrt2 \left [ \sqrt{(x^2+2x+1+y^2 ) } \right ]=0$
1+y = 0
y = -1
$x+\sqrt2 \sqrt{x^2+2x+2}=0$
$\sqrt2 \sqrt{x^2+2x+2}=-x$
$2x^2+4x+4=x^2$
$x^2+4x+4=0$
$(x+2)^2=0$
$x=-2$
Hence,$z=x+iy=-2-i$

Question:23

Write the complex number $z=\frac{1-i}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$ in polar form

Answer:

$z=\frac{1-i}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\frac{\sqrt{2}\left [ \frac{1}{\sqrt{2}}-i1/\sqrt{2} \right ]}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\frac{\sqrt{2}\left [\cos\frac{\pi}{4}-isin\frac{\pi}{4} \right ]}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\sqrt2\left [ \cos \left ( -\frac{\pi}{4}-\frac{\pi}{3}\right )+i\sin\left (-\frac{\pi}{4}-\frac{\pi}{3} \right ) \right ]$
$=\sqrt2\left [ \cos \left ( -\frac{7\pi}{12}\right )+i\sin\left (-\frac{7\pi}{12} \right ) \right ]$
$=-\sqrt2\left [ \cos \left ( \frac{5\pi}{12}\right )+i\sin\left (\frac{5\pi}{12} \right ) \right ]$

Question:24

If z and w are two complex numbers such that $|zw| = 1$ and $arg(z) - arg (w) = \pi/2$, then show that $zw=-i$.

Answer:

Let $z=\left | z \right |(\cos\theta_1 + i sin\theta_1 )$ and $w=\left | w \right |(\cos\theta_2 + i sin\theta_2 )$
$|zw|=|z||w|=1$
Also $arg(z)-arg(w)= \frac{\pi}{2}$
$\theta_1-\theta_2=\frac{\pi}{2}$
$\bar{z} w=|z|\left (\cos \theta _{1}-i\sin\theta _{1} \right )w$
$w=|w|\left (\cos \theta _{2}+i\sin\theta _{2} \right )=1$
$\bar{z}w$$=|z||w|(\cos (-\theta_1)+i sin(-\theta_1 ))(cos\theta_2+i \sin\theta_2 )$
$\bar{z}w$$=\cos [(\theta _2 - \theta _1 )+isin( \theta _2- \theta _1 )]$
$=\left [ \cos \left ( -\frac{\pi}{2} \right ) +i\sin\left ( -\frac{\pi}{2} \right )\right ]$
$=1[0-i]=-i$

Question:25

i) For any two complex numbers z₁, z₂ and any real numbers a, b, $|az_1 - bz_2|^2 + |bz_1 + az_2|^2 =....$

ii) The value of $\sqrt{-25} * \sqrt{-9}$ is

iii) The number $\frac{(1-i)^3}{1-i^3}$ is equal to .......

iv) The sum of the series $i+i^2+i^3+...+i^{1000}$ upto 1000 terms is ...

v) Multiplicative inverse of 1 + i is ................

vi) If $z_1 \ and \ z_2$ are complex numbers such that $z_1 + z_2$ is a real number, then $z_2$ =

vii) $arg (z)+arg (\bar{z} )$ $(\bar{z}\neq0)$ is ....

viii) If $|z+4|\leq 3$then the greatest and least values of |z+1| are ..... and .....

ix) if $\frac{z-2}{z+2}=\frac{\pi}{6}$then the locus of z is .......

x) If $|z|=4$ and $arg (z) = \frac{5\pi}{6}$ then z=

Answer:

i) $|az_1 - bz_2|^2 + |bz_1 + az_2|^2$
$=|az_1 |^2+|bz_2 |^2-2Re(az_1.b(\bar{z_2} ) +|bz_1 |^2+|az_2 |^2+2Re(az_1.b\bar{z_2} ) ?)$
$=|az_1 |^2+|bz_2 |^2+|bz_1 |^2+|az_2 |^2=(a^2+b^2 )(|z_1 |^2+|z_2 |^2 )$
ii)$\sqrt{-25}*\sqrt{-9}=5i*3i=15i^2=-15$
iii) $\frac{(1-i)^3}{1-i^3}=\frac{(1-i)^3}{(1-i)(1+i+i^2 )}$
$=\frac{(1-i)^3}{(1+i-1) }= \frac{1+i^2-2i}{}i$
$=\frac{1-1-2i}{i}=-\frac{2i}{i}=-2$
iv) $i+i^2+i^3+...+i^{1000}=0 \left [ \sum_{n=1}^{1000}i^n=0 \right ]$
v) $\frac{1}{1+i}=1*\frac{1-i}{(1+i)(1-i)} =\frac{1}{2} (1-i)$
vi) Let $z_1=x_1+iy_1 \: \: and \: \: z_2=x_2+iy_2$
$z_1+z_2=(x_1+x_2 )+i(y_1+y_2 )$
If $z_1+z_2$ is real then $y_1+y_2=0$
$y_1=-y_2$
$z_2=x_2-iy_1$
$z_2=x_1-iy_1 (\: \: when\: \: x_1=x_2 )$
So, $z_2=\bar{z_1}$
vii) $arg (z)+arg (\bar{z} )$
$If\: \: arg (z)= \theta , then \arg (\bar{z})=-\theta$
$\theta+(-\theta)=0$
viii) $|z+4|\leq 3$
$=|z+4-3|\leq|z+4|+|-3|$
$=|z+4-3|\leq3+3$
$=|z+4-3|\leq6$
The gratest value is 6 and the least value of $|z+1|$ is 0
ix)$\frac{z-2}{z+2}=\frac{\pi}{6}$
Let $z=x+iy$
$\left |\frac{x+iy-2}{x+iy+2} \right |=\left |\frac{(x-2)+iy}{(x+2)+iy} \right |= \frac{\pi}{6}$
$6|(x-2)+iy|=\pi|(x+2)+iy|$
$6\sqrt{(x-2)^2+y^2 }=\pi\sqrt{(x+2)^2+y^2 }$
$36[x^2+4-4x+y^2 ]=\pi ^2[x^2+4+4x+y^2 ]$
$36x^2+144-144x+36y^2=\pi^2 x^2+4\pi^2+4\pi^2 x+\pi^2 y^2$
$(36-\pi^2 ) x^2+(36-\pi^2 )-(144+4\pi^2 )x+144-4\pi^2=0$
which represents an equation of a circle
x) $|z|=4$ and $arg (z) = \frac{5\pi}{6}$
Let $z=x+iy$
$|z|= \sqrt{x^2+y^2} =4$
$x^2+y^2=16$
$arg(z)=\tan^{-1}\frac{y}{x}=\frac{5\pi}{6}$
$\frac{y}{x}=\tan\left ( \pi-\frac{\pi}{6} \right )$
$=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}$
$x=-\sqrt3 y$
$(-\sqrt3 y)^2+y^2=16$
$3y^2+y^2=16$
$4y^2=16$
$y^2=4$
$y=\pm2$
$x=-2\sqrt3$
$z=-2\sqrt3+2i$

Question:26

State True of False for the following:

i) The order relation is defined on the set of complex numbers.
ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction
iii) For any complex number z, the minimum value of |z| + |z – 11 is 1
iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1)
v) If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im (z²) = 0
vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z₁ and Z₂ be two complex numbers such that |z, + z₂| = |z₁ j + |z₂|, then arg (z₁ – z₂) = 0.
(viii) 2 is not a complex number.

Answer:

(i) Comparison of two purely imaginary complex numbers is not possible. However, the two purely real complex number can be compared. So, it is false.

(ii) Let $z= x+iy$

$z.i = (x+iy)i =xi= -y$ which rotates at angle of 180. So, it is ‘false’.

(iii) Let $z= x+iy$

$|z|+|z-1|= \sqrt{x^2+y^2} +\sqrt{(x-1)^2+y^2}$

The value of $|z|+|z-1|$ is minimum when $x=0,y=0 \, \: \: i.e.,1$

Hence, it is true.

iv) Let $z= x+iy$

Given that $\left | z-1 \right |=\left |z-i \right |$

$|x+iy-1|=|x+iy-i|$

$|(x-1)+iy|=|x-(1-y)i|$

$\sqrt{(x-1)^2+y^2}=\sqrt{x^2+(1-y)}^2$

$(x-1)^2+y^2=x^2+(1-y)^2$

$x^2-2x+1+y^2=x^2+1+y^2-2y$

$-2x+2y=0 x-y=0$ which is a straight line slope=1

Now, equation of line through the point 1,0and 0,1

$y-0=\frac{1-0}{0-1} (x-1)$

$y=-x+1$ whose slope=-1

Multiplication of the slopes of two lines =-1*1=-1

So, they are perpendicular. Hence, it is true.

v)Let $z= x+iy$ $z\neq0 \: \: and \: \: Re(z)=0$

Since, real part is 0 $x=0, z=0+iy=iy$

$1m(z^2 )=y^2 i^2=-y^2$which is real Hence, it is False.

vi) $(z-4)<|z-2|$

Let $z= x+iy$

$|x+iy-4|<|x+iy-2|$

$|(x-4)+iy|<|(x-2)+iy|$

$\sqrt{(x-4)^2+y^2 }<\sqrt{(x-2)^2+y^2}$

$(x-4)^2+y^2<(x-2)^2+y^2$

$(x-4)^2<(x-2)$

$x^2+16-8x<x^2+4-4x$

$8x+4x<-16+4$

$-4x<-12$

$x>3$ Hence, it is true.

vii) Let $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$

$|z_1+z_2 |=|z_1 |+|z_2 |$

$|x_1+iy_1+x_2+iy_2 |=|(x_1+iy_1 )+(x_2+y_2 i)|$

$= \sqrt{(x_1+x_2 )^2+(y_1+y_2 )^2 }$

$=\sqrt{x_1^2+y_1^2} +\sqrt{x_2^2+y_2^2 }$

Squaring both sides, we get $(x_1^2+x_2^2+2x_1 x_2+y_1^2+y_2^2+2y_1 y_2$)

$=x_1^2+y_1^2+x_2^2+y_2^2+2\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

$=2x_1 x_2+2y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

$=x_1 x_2+y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$

Again squaring on both sides we get $x_1^2 x_2^2+x_1^2 y_2^2+2x_1 y_1 x_2 y_2=x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 2x_1 y_1 x_2 y_2=x_1^2 y_2^2+x_2^2 y_1^2$

$x_1^2 y_1^2+x_2^2 y_1^2-2x_1 y_1 x_2 y_2=0$

$(x_1 y_2-x_2 y_1 )^2=0$

$x_1 y_2-x_2 y_1=0$

$x_1 y_2=x_2 y_1$

$\frac{x_1}{y_1}=\frac{x_2}{y_2}$

$\frac{y_1}{x_1}=\frac{y_2}{x_2}$

$arg(z_1)=arg(z_2)=0$ Hence, it is false.

(viii) Since, every real number is a complex number. So, 2 is a complex number. Hence, it is false.

Question:27

Answer:

a)Given $z=i+\sqrt3$

Polar form of z $=r[cos\theta+isin\theta]=i+\sqrt3$.

$r=\sqrt{3+1} =2$

And $\tan \alpha = \frac{1}{\sqrt3}$

$\alpha = \frac{\pi}{6}$

Since $x>0,y>0$

Polar form of z $=2[cos\frac{\pi}{6}+isin\frac{\pi}{6}]$

b) Given that $z=-1+\sqrt3=-1+\sqrt3 i$

Here argument z $=\tan^{-1}\left | \frac{\sqrt3}{-1} \right |= \tan^{-1}\sqrt{3}= \frac{\pi}{3}$

So, $\alpha = \frac{\pi}{3}$

Since, x<0 and y>0 $\theta=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

c) $|z+2|=|z-2|$

$|x+iy+2|=|x+iy-2|$

$|(x-2)+iy|=|(x-2)+iy|$

$\sqrt{(x+2)^2+y^2} =\sqrt{(x-2)^2+y^2}$

$(x+2)^2+y^2=(x-2)^2+y^2$

$(x+2)^2=(x-2)^2$

$x^2+4+4x=x^2+4-4x$

$8x=0 , x=0$

which represents equation of-y axis and is perpendicular to the line joining the points (-2,0) and (2,0)

d) $|z+2i|=|z-2i|$

$(x+iy+2i)=|x+iy-2i|$

$|x+(y+2)i|=|x+(y-2)i|$

$\sqrt{x^2+(y+2)^2} =\sqrt{x^2+(y-2)^2 }$

$x^2+(y+2)^2=x^2+(y-2)^2$

$(y+2)^2=(y-2)^2$

$y^2+4+4y=y^2+4-4y$

$8y=0 ,y=0$

which is the equation of x-axis and is perpendicular to the line segment joining (0,-2)and (0,2)

e) $|z+4i|\geq3$

$|x+iy+4|\geq3$

$|x+(y+4)i|\geq3$

$\sqrt{x^2+(y+4)^2} \geq3$

$x^2+(y+4)^2\geq9$

$x^2+y^2+8y+16 \geq9$

$x^2+y^2+8y+7 \geq0$ $r=\sqrt{(4)^2-7}=3$

which represents a circle on or outside having centre (0,-4)

f) $|z+4|\leq3$

Let $z=x+iy$ Then, $|x+iy+4|\leq3$

$|(x+4)+iy|\leq3$

$\sqrt{(x+4)^2+y^2 } \leq3$

$x^2+8x+16+y^2\leq9$

$x^2+y^2+8x+7 \leq0$ which is a circle having centre (-4,0) and $r=\sqrt{(4)^2-7}=\sqrt9=3$ and is on the circle.

g) Let $z=\frac{1+2i}{1-i}$

$\frac{1+2i}{1-i}* \frac{1+i}{1+i}=\frac{1+i+2i+2i^2}{1-i^2}$

$=\frac{1+i+2i-2}{1+1}= \frac{-1+3i}{2}$

$=\frac{1+i+2i-2}{1+1}= \frac{-1+3i}{2}$

$=-\frac{1}{2}+\frac{3}{2}i$ which lies in third quadrant..

h) Given that z=1-i

Reciprocal of $z=\frac{1}{z}=\frac{1}{1-i}*\frac{1+i}{1+i}$

$=\frac{1+i}{1+i^2}$ which lies in first quadrant.

$So,(a)-(v),(b)-(iii),(c)-(i),(d)-(iv),(e)-(ii),(f)-(vi),(g)-(viii),(h)-(vii)$

Question:28

What is the conjugate of $\frac{2-i}{\left (1-2i \right )^2}$?

Answer:

$\frac{2-i}{\left (1-2i \right )^2}$$= \frac{2-i}{1+4i^2-4i}$
$= \frac{2-i}{-3-4i}= \frac{2-i}{-3-4i}*\frac{-3+4i}{-3+4i}$
$=\frac{-6+8i+3i-4i^2}{9-16i^2}$
$\frac{-6+11i+4}{9-16i^2}=\frac{-2+11i}{9+16}$
$\frac{-2+11i}{25}=\frac{-2}{25}+\frac{11}{25}i$
Hence, $\bar{z}=-\frac{2}{25}-\frac{11}{25}i$

Question:29

If $|z_1| = |z_2|$, is it necessary that $z_1 = z_2$?

Answer:

Let $z_1=x_1+iy_1 \: \: and\: \: z_2=x_2+iy_2$
$|x_1+iy_1 |=|x_2+iy_2 |$
$= \sqrt{x_1^2+y_1^2} = \sqrt{x_2^2+y_2^2}$
$x_1^2+y_1^2=x_2^2+y_2^2$
$\\$Consider $ z_1 = 3 + 4i \\ z_2 = 4 + 3i$
$z_1\neq z_2$ Hence, it is not necessary that $z_1= z_2$

Question:31

Find z if $|z| = 4$ and $arg (z) = 5\pi/6$..

Answer:

Polar form of $z=r[cos \theta+i\sin \theta]$
$=4\left [ \cos \frac{5\pi}{6}+i\sin \frac{5\pi}{6} \right ]$
$=4\left [ \cos\left ( \pi-\frac{\pi}{6} \right )+i\sin \left ( \pi-\frac{\pi}{6} \right ) \right ]$
$=4\left [ -\cos\left (\frac{\pi}{6} \right )+i\sin \left (\frac{\pi}{6} \right ) \right ]$
$=4\left [-\frac{\sqrt3}{2}+i\frac{1}{2} \right ]$
$=-2\sqrt3+2i$

Question:32

Find $(1+i)\frac{(2+i)}{(3+i)}$.

Answer:

$\left | (1+i)\frac{(2+i)}{(3+i)}*\frac{3-i}{3-i} \right |=\left | (1+i).\frac{6-2i+3i-i^2}{9-i^2} \right |$
$=\left | \frac{(1+i)(7+i)}{9+1} \right |$
$=\left | \frac{7+i+7i+i^2}{9+1} \right |$
$=\left | \frac{7+i+7i+i^2}{10} \right |$
$=\left | \frac{7+8i-1}{10} \right |$
$=\left | \frac{6+8i}{10} \right |$
$=\left |\frac{3}{5}+\frac{4}{5}i \right |$
$=\sqrt{\left (\frac{3}{5} \right )^2+\left (\frac{4}{5} \right )^2}=1$

Question:33

Find principal argument of $(1+i\sqrt3)^2$.

Answer:

$\begin{array}{c} (1+i \sqrt{3})^{2}=1+i^{2} \cdot 3+2 \sqrt{3} i \\ =1-3+2 \sqrt{3} i=-2+2 \sqrt{3} i \\ \tan \alpha=\left|\frac{2 \sqrt{3}}{-2}\right| \\ \tan \alpha=|-\sqrt{3}|=\sqrt{3} \\ \tan \alpha=\tan \frac{\pi}{3} \\ \alpha=\frac{\pi}{3} \end{array}$

Now Re(z)<0 and image(z)>0 $arg(z)=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$
$\text {Hence, the principal arg}=\frac{2 \pi}{3}$

Question:34

Where does z lie, if $|\frac{z-5i}{z+5i}|=1$.

Answer:

$\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{x+y i-5 i}{x+i y+5 i}\right|=1 \\ |x+(y-5) i|=|x+(y+5) i| \\ x^{2}+(y-5)^{2}=x^{2}+(y+5)^{2} \\ (y-5)^{2}=(y+5)^{2} \\ \qquad \begin{array}{r} y^{2}+25-10 y=y^{2}+25+10 y \\ 20 y=0 \quad y=0 \end{array} \end{array}$

Hence, z lies on x-axis i.e., real axis.

Question:35

$\sin x + i \cos 2x$ and $\cos x - i \sin 2x$ are conjugate to each other for:
A. $x = n\pi$

B. $x = \left ( n+\frac{1}{2} \right )\frac{\pi}{2}$
C. $x = 0$
D. No value of x

Answer:

The answer is the option (c).

Comparing the real and imaginary part, we get

$\tan \mathrm{x}=1 \text { and } \tan 2 \mathrm{x}=1$
$\tan 2 \mathrm{x}= \frac{2\tan x }{1-\tan^2 x}=1$
The above value is not satisfied by tan x = 1. Hence no value of x is possible.

Question:36

The real value of α for which the expression $\frac{\left ( 1-i\sin\alpha \right )}{1+2i\sin\alpha}$ is purely real is:
A.$\left ( n+1 \right )\frac{\pi}{2}$
B.$\left ( 2n+1 \right )\frac{\pi}{2}$
C.$n\pi$
D. None of these

Answer:

The answer is the option (c).
Let $z=\frac{\left ( 1-i\sin\alpha \right )}{1+2i\sin\alpha}$
$\begin{aligned} &\begin{array}{c} =\frac{(1-i\sin \alpha)(1-2 i \sin \alpha)}{(1+2 i \sin \alpha)(1-2 i \sin \alpha)} \\\\ =\frac{1-2 i \sin \alpha-i \sin \alpha+2 i^{2} \sin ^{2} \alpha}{(1)^{2}-(2 i \sin \alpha)^{2}} \\\\ =\frac{1-3 i \sin \alpha-2 \sin ^{2} \alpha}{1-4 i^{2} \sin ^{2} \alpha}=\frac{\left(1-2 \sin ^{2} \alpha\right)-3 i \sin \alpha}{1+4 \sin ^{2} \alpha} \\\\ \frac{-3 i \sin \alpha}{1+4 \sin ^{2} \alpha}=0 \\\\ \sin \alpha=0 \quad \operatorname{} \alpha=n \pi \end{array}\\ &\text { Hence, c is correct. } \end{aligned}$

Question:37

If z = x + iy lies in the third quadrant, the $\frac{\bar{z}}{z}$also lies in the third quadrant if
A. x > y > 0
B. x < y < 0
C. y < x < 0
D. y > x > 0

Answer:

The answer is the option (b).
If z lies in the third quadrant,So,$x<0 \: \: and\: \: y<0\: \: , z=x+iy$
$\begin{aligned} \frac{\bar{z}}{z} &=\frac{x-i y}{x+i y}=\frac{x-i y}{x+i y} * \frac{x-i y}{x-i y} \\ &=\frac{x^{2}+i^{2} y^{2}-2 x y i}{x^{2}-i^{2} y^{2}} \\ &=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}-\frac{2 x y}{x^{2}+y^{2}} i \end{aligned}$
When z lies in third quadrant then $\frac{\bar{z}}{z}$ will also be in third quadrant.
$\begin{array}{c} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}<0 \text { and } \frac{2 x y}{x^{2}+y^{2}}>0 \\\\ x^{2}-y^{2}<0 \text { and } 2 x y>0 \\\\ x^{2}<y^{2} \text { and } x y>0 \\\\ \text { So, } x<y<0 \end{array}$
Hence, b is correct..

Question:38

The value of $(z + 3) (\bar{z}+3)$ is equivalent to
A.$|z + 3|^2$
B. |z – 3|
C.$z^2 + 3$
D. None of these

Answer:

The answer is the option (a).
$Let\: \: z=x+iy \: \: So,\: \: (z+3)(\bar{z} +3)=(x+iy+3)(x-iy+3)$
$=\left [ \left ( x+3 \right ) +iy\right ]\left [ \left ( x+3 \right ) -iy\right ]$
$=(x+3)^2-y^2 i^2=(x+3)^2+y^2$
$=|x+3+iy|^2=|z+3|^2$

Question:39

If $\left ( \frac{1+i}{1-i} \right )^{x}=1$ , then
A. $x = 2n + 1$
B. $x = 4n$
C. $x = 2n$
D.$x = 4n + 1$

Answer:

The answer is the option(b).
$\begin{array}{c} \left(\frac{1+i}{1-i}\right)^{x}=1 \\\\ \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{x}=1 \\\\ \left(\frac{1-1+2 i}{1+1}\right)^{x}=1 \\\\ \left(\frac{2 i}{2}\right)^{x}=1(i)^{x}=1 \end{array}$
$x = 4 n, n \epsilon N$
Hence, the correct option is

A real value of x satisfies the equation $\frac{3-4 i x}{3+4 i x}=\alpha-i \beta$ ($\alpha, \beta \in R$) , if $\alpha ^2 +\beta ^2 =$
A. 1
B. –1
C. 2
D. –2

Answer:

The answer is the option (a).

$\begin{aligned} &\text {Given that} \frac{3-4 i x}{3+4 i x}=\alpha-i \beta\\ &=\frac{3-4 i x}{3+4 i x} * \frac{3-4 i x}{3-4 i x}=\alpha-i \beta\\ &\frac{9-12 i x-12 i x+16 i^{2} x^{2}}{9-16 i^{2} x^{2}}=\alpha-i \beta\\ &=\frac{9-16 x^{2}}{9+16 x^{2}}-\frac{24 x}{9+16 x^{2}}=\alpha-i \beta \ldots \ldots .(i)\\ &\frac{9-16 x^{2}}{9+16 x^{2}}+\frac{24 x}{9+16 x^{2}}=\alpha+i \beta \ldots \ldots \ldots(ii)\\\end{aligned}$
Multiplying eqn (i )and (ii )we get
$\left (\frac{9-16 x^{2}}{9+16 x^{2}} \right )^{2}+\left (\frac{24 x}{9+16 x^{2}} \right )^{2}=\alpha^{2}+ \beta^{2}$
$\frac{81+256 x^{4}-288 x^{2}+576x^2}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}$
$\begin{aligned}\frac{81+256 x^{4}+288 x^{2}}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}\\ \frac{\left(9+16 x^{2}\right)^{2}}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}=1 \end{aligned}$

Question:41

Which of the following is correct for any two complex numbers $z_1\: \: and\: \: \: \: z_2$ ?
A. $|z_1 z_2| = |z_1||z_2|$
B. $arg (z_1z_2) = arg (z_1). Arg (z_2)$
C. $|z_1 + z_2| = |z_1|+ |z_2|$
D. $|z_1 + z_2| \geq |z_1| - |z_2|$

Answer:

The answer is the option (a).
$\begin{array}{c} z_{1}=r_{1}\left(\cos \theta_{1}+i \sin \theta_{2}\right) \\ \left|z_{1}\right|=r_{1} \\ z_{2}=r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \left|z_{2}\right|=r_{2} \\\\ z_1z_2=r_{1} r_{2}\left[\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right] \\ =r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right) \cdot r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \end{array}$
$\begin{array}{c} =r_{1} r_{2}\left[\left(\cos \theta_{1} \cos \theta_{2}-\sin \theta_{1} \sin \theta_{2}\right)+i\left(\sin \theta_{1} \cos \theta_{2}+\cos \theta_{1} \sin \theta_{2}\right)\right] \\ =r_{1} r_{2}\left[\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right]\left|z_{1} z_{2}\right|=\left|z_{1} \| z_{2}\right| \end{array}$

Hence, option a is correct.

Question:42

The point represented by the complex number $2 -i$ is rotated about origin through an angle $\pi/2$ in the clockwise direction, the new position of point is:
A. $1 + 2i$
B. $-1-2i$
C. $2 + i$
D. $-1 + 2i$

Answer:

The answer is the option (b).
If z rotated through an angle of $\pi/2$about the origin in clockwise direction .
Then the new position $=z.e^{-\left ( \frac{\pi}{2} \right )}$
$=\left (2-i \right ).e^{-\left ( \frac{\pi}{2} \right )}$
$=\left (2-i \right ).\left [ \cos\left (- \frac{\pi}{2} \right )+isin\left (- \frac{\pi}{2} \right ) \right ]$
$=\left (2-i \right ).\left (0-i \right )$
$=-1-2i$
Hence, the correct option is (b)

Question:43

Let $x, y \in R$, then $x + iy$ is a non-real complex number if:
A. $x = 0$
B.$y = 0$
C. $x \neq 0$
D. $y \neq 0$

Answer:

The answer is the option (d).
$x+iy$ is a non real complex number if $y\neq0$,

Hence, d is correct

Question:44

If $a + ib = c + id$, then
A. $a^2 + c^2 = 0$
B. $b^2 + c^2 = 0$
C. $b^2 + d^2 = 0$
D. $a^2 + b^2 = c^2 + d^2$

Answer:

The answer is the option (d).
$a+ib=c+id$
$|a+ib|=|c+id|$
$\sqrt{a^2+b^2} =\sqrt{c^2+d^2}$
Squaring both sides, $a^2+b^2=c^2+d^2$
Hence, d is correct

Question:45

The complex number z which satisfies the condition $\left |\frac{i+z}{i-z} \right |=1$ lies on
A. circle $x^2 + y^2 = 1$
B. the x-axis
C. the y-axis
D. the line $x + y = 1$

Answer:

The answer is the option (b).
$\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{i+x+i y}{i-x-y i}\right|=1 \\\\ \left|\frac{x-(y+1) i}{-x-(y-1) i}\right|=1 \\\\ |x+(y+1) i|=|-x-(y-1) i| \\\\ \sqrt{x^{2}+(y+1)^{2}}=\sqrt{x^{2}+(y-1)^{2}} \\\\ x^{2}+(y+1)^{2}=x^{2}+(y-1)^{2} \\\\ y^{2}+2 y+1=y^{2}-2 y+1 \\ 2 y=-2 y \\\\ y=0 ; x-a x i s \end{array}$

Hence, b is correct

Question:46

If z is a complex number, then
A.$|z^2| > |z|^2$
B. $|z^2| \: \: equals\: \: |z|^2$
C.$|z^2| < |z|^2$
D. $|z^2| \geq |z|^2$

Answer:

. The answer is the option (b).
$Let \: \: z=x+iy \: \: \: \: ,|z|^2=x^2+y^2$
$Now,z^2=x^2+y^2 i^2+2xyi$
$z^2=x^2-y^2+2xyi$
$|z|^2=\sqrt{(x^2-y^2 )^2+(2xy)^2 }$
$=\sqrt{x^4+y^4+2x^2 y^2} =\sqrt{(x^2+y^2 )^2}$
$|z|^2=x^2+y^2=|z|^2$

Hence, b is correct

Question:47

$|z_1 + z_2| = |z_1| + |z_2|$ is possible if
A. $z_1 =\bar{z_1}$

B. z₂₌z₁
C. $arg (z_1) = arg (z_2)$
D. $|z_1| = |z_2|$

Answer:

The answer is the option (c).
Since, $|z_1+z_2 |=|z_1 |+|z_2 |$
$|z_1+z_2 |=|r_1 \left ( \cos \theta _1+i\sin\theta _1 \right )+r_2 \left ( \cos \theta _2+i\sin\theta _2 \right )|$
$=\sqrt{r_{1}^{2} \cos ^{2} \theta_{1}+r_{2}^{2} \cos ^{2} \theta_{2}+2 r_{1} r_{2} \cos \theta_{1} \cos \theta_{2}+r_{1}^{2} \sin ^{2} \theta_{1}+r_{2}^{2} \sin ^{2} \theta_{2}+2 r_{1} r_{2} \sin \theta_{1} \sin \theta_{2}}$
So, $\begin{array}{c} =\sqrt{r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)} \\ \left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right| \\\\ \sqrt{r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)}=r_{1}+r_{2} \end{array}$
Squaring both sides,
$\begin{array}{c} r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \\ 2 r_{1} r_{2}-2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=0 \\ 1-\cos \left(\theta_{1}-\theta_{2}\right)=0 \\ \cos \left(\theta_{1}-\theta_{2}\right)=1 \\ \theta_{1}-\theta_{2}=0 \\ \theta_{1}=\theta_{2} \\ \operatorname{so} \arg \left(z_{1}\right)=\arg \left(z_{2}\right) \end{array}$

Question:48

The real value of θ for which the expression $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is a real number is:
A. $n\pi+\frac{\pi}{4}$

B. $n\pi+(-1)^n\frac{\pi}{4}$
C. $2n\pi+\frac{\pi}{2}$
D. none of these

Answer:

On solving we get
$\frac{1+3 i \cos \theta-2 \cos ^{2} \theta}{1-4 i^{2} \cos ^{2} \theta} \\=\frac{1+3 i \cos \theta-2 \cos ^{2} \theta}{1+4 \cos ^{2} \theta}$
$=\frac{\left(1-2 \cos ^{2} \theta\right)+3 i \cos \theta}{1+4 \cos ^{2} \theta}$
If z is real number then
$\frac{3 \cos \theta}{1+4 \cos ^{2} \theta}=0$
$\\3 \cos \theta=0 \\\cos \theta=0$

Question:49

The value of $arg (x)$ when $x < 0$is:
A. 0
B. $\frac{\pi}{2}$
C. π
D. none of these

Answer:

The answer is the option (c).
$\\ Let\: \: z= -x+0i \: \: and \: \: x<0 \\ |z|=\sqrt{(-1)^2+(0)^2 }=1,x<0$

Since, the point (-x,0)lies on the negative side of the real axis , principal argument (z)=π

Hence, option c is correct

Question:50

If $f(z)=\frac{7-z}{1-z^2}$where $z = 1 + 2i$, then $|f(z)|$ is
A. $\frac{\left | z \right |}{2}$

B. $|z|$
C. $2|z|$
D. none of these

Answer:

The answer is the option (a).
Given that $\\ z=1+2i \\ |z|=\sqrt{(1)^2+(2)^2 }=\sqrt5$
Now $f(z)=\frac{7-z}{1-z^2}$
$\begin{aligned} =\frac{7-(1+2 i)}{1-(1+2 i)^{2}} &=\frac{7-1-2 i}{1-1-4 i^{2}-4 i} \\ =\frac{6-2 i}{4-4 i} &=\frac{3-i}{2-2 i} \end{aligned}$
$\begin{aligned} =\frac{3-i}{2-2 i} * \frac{2+2 i}{2+2 i}=\frac{6+6 i-2 i-2 i^{2}}{4-4 i} \\ =\frac{6+4 i+2}{4+4} &=\frac{8+4 i}{8} \\ =1+\frac{1}{2} i & \end{aligned}$
$f(z)=\sqrt{(1)^{2}+\left(\frac{1}{2}\right)^{2}}=\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}=\frac{|z|}{2}$

Hence, option a is correct